PHYS 132, WEEK 7, CH 16
16.9.
Solve:
Because the atomic mass number of gold is 197, 1.0 mole of gold has a mass of 197 g or 0.197 kg. The volume of 1.0 mole
of gold is
V
=
L
3
=
M
r
Au
L
=
0.197 kg
(
)
19, 300 kg m
3
1 3
=
2.17 cm
16.12.
Solve:
Let
T
F
=
T
C
=
T
:
T
F
=
9
5
T
C
+
32
°
T
=
9
5
T
+
32
°
T
= 
40
°
That is, the Fahrenheit and the Celsius scales give the same numerical value at

40
°
.
16.19.
Model:
Treat the gas in the sealed container as an ideal gas.
Solve:
(a)
From the ideal gas law equation
pV
=
nRT
, the volume
V
of the container is
V
=
nRT
p
=
2.0 mol
(
)
8.31 J mol K
(
)
273
+
30
(
)
K
[
]
1.013
10
5
Pa
=
0.0497 m
3
Note that pressure
must
be in Pa in the ideal gas law.
(b)
The beforeandafter relationship of an ideal gas in a sealed container (constant volume) is
p
1
V
T
1
=
p
2
V
T
2
p
2
=
p
1
T
2
T
1
=
1.0 atm
(
)
273
+
130
(
)
K
273
+
30
(
)
K
=
1.33 atm
Note that gaslaw calculations
must
use T in kelvins.
16.24.
Model:
Treat the oxygen gas in the cylinder as an ideal gas.
Solve:
(a)
The number of moles of oxygen is
n
=
M
M
mol
=
50 g
32 g mol
=
1.563 mol
(b)
The number of molecules is
N
=
nN
A
=
1.563 mol
(
)
6.02
10
23
mol

1
(
)
=
9.41
10
23
.
(c)
The volume of the cylinder
V
=
p
r
2
L
=
p
0.10 m
(
)
2
0.40 m
(
)
=
1.257
10

2
m
3
.
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 Spring '08
 Sharpe
 Mass, mol

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